Economics When Applying Shunt Capacitors · All these benefits will have financial benefits to the...

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Economics When Applying Shunt Capacitors

Transcript of Economics When Applying Shunt Capacitors · All these benefits will have financial benefits to the...

Page 1: Economics When Applying Shunt Capacitors · All these benefits will have financial benefits to the utility, or whoever provides the capacitor bank on their system. Consolidation of

Economics When Applying Shunt Capacitors

Page 2: Economics When Applying Shunt Capacitors · All these benefits will have financial benefits to the utility, or whoever provides the capacitor bank on their system. Consolidation of

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Economics When Applying Shunt Capacitors Managing your electrical loads offers many challenges. Your plant needs all the power it can muster while conserving as much energy as possible and yet maintain an efficient operation. Looking at the power consumed it is necessary to supply two types of powers. The first is "Active Power" (Kilowatts); the second is “Reactive Power” (KVAR). The active power is supplied by the utility incoming. The reactive power can be furnished by the system, or by the use of static shunt capacitors. It has been established that shunt power capacitors are the most economical source for the reactive power (KVAR) required by the loads and lines when operating at less than unity power factor (100%). Studies show that supply systems from the power companies lines require reactive power (KVAR) in addition to the consumers electric load. Looking at a simple system, if the only reactive power source is the central station generation, this reactive power will have to be generated by the generators and than transmitted over the lines to the loads. If the power company has capacitor banks at the supply substation this will aid in supplying reactive power. However, if the consumer’s plant has a poor power factor the substation and the lines to the plant will have to handle the additional currents required. Coupled with the additional currents developed by poor power factor, there is also corresponding power loss (I2R loss) associated with transmission and distribution of reactive power current to the plants load. These losses create an undesirable voltage reduction on the lines to the plant. Shunt capacitors affect the voltage rise when connected to the system. The addition of switched capacitors not only improves the voltage levels, but also provides an effective method of controlling the voltage levels. •The installation of power capacitors enables a utility, as will as the industrial customer, to realize savings on their systems. The following benefits can be realized. •Raised Voltage Levels •Released Generation Capacity •Released system capacity •Reductions of System Loses •Regulations of Voltage Levels The utility can witness benefits on their generation, EHV transmission, sub-transmission and distribution systems with power capacitors. Voltage Considerations Voltage Drop: Simply, the voltage drop is basic and is due to the impedance in the line. The impedance consists of resistance, which creates IR voltage drop and reactance, which creates IXL voltage drop. The combination of these two drops is known as the impedance drop, or IZ drop.

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Figure 1

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Table 1

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Table 2

Percent Voltage Drop:

%Voltage Drop =

D =Line length KV = phase to phase voltage

Let

KW = 1000 Pf = 85% R = 0.699/mile XL = 0.712 Ohms D = 10 Miles

Then

KV = 12.47 ARCCOS 0.85 = 31.80

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SIN = 0.527

With a 7.33% voltage drop the total voltage drop will be 914.57 volts. The line voltage at this point will be 11,555.43 volts. If the secondary voltage is 480 volts normally this voltage drop will be 444.8 volts. Voltage Rise:

Figure 2

D =Line length KV = phase to phase voltage from above KV = 12.47 KW = 1000 Pf = 85% XL = 0.712 Ohms D = 10 Miles

To correct the power factor to approximately 95%, we will use 300 KVAR. This will correct this system to approximately 97%

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Then

With a 1.374% voltage rise the voltage at the load VL will be increased by 159 volts. The line voltage at this point will be 11,714.2 volts. With the secondary voltage of 444.8 volts, voltage will raise to 461.44 volts. Percentage Voltage Drop In a Transformer The percentage voltage drop may be calculated by the following:

Voltage Rise Through Transformers The above discussion shows the voltage drop and rise on the power line and drop in the transformers. Every transformer will also experience a voltage rise from generating source to the capacitors. This rise is independent of load or power factor and may be determined as follows:

KVAR =Applied Kilo-vars KVA = KVA of the transformer Xt = Transformer Reactance in %Using the 300 Kvar bank given above and assuming a 1200 KVA transformer with 5.75% reactance we would have:

New Total Voltage Improved Drop New net Voltage Drop =Total Drop – Total Rise Using the values from above, the new improved voltage drop will be:

Voltage Drop in Line =7.3300%

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Voltage Drop in the Transformer = 0.7480%

Total Voltage Drop =8.0780% Voltage Rise in Line =1.3740%

Voltage Rise in the Transformer = 1.4375%

Total Voltage Rise =2.8115% Net Voltage Drop =8.078% - 2.8115% = 5.2665%

Substation Capacity Released Assuming the 1200 KVA transformer is in the customers substation and is used with the other values used in the equations above the released substation capacity is as follows:

A quick approximation of KVA increase can be as follows:

Generating Capacity Released To determine this capacity released the formula used for the Substation Capacity Release can also be used. We would replace the substation transformer (KVAS) with the generating capacity (KVAG). Assuming the KVAG is 20MVA we have the following:

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Or, by the simplified formula

Increased Feeder Capacity Feeder capacity is limited basically by permissible voltage drop rather than thermal conditions. The application of capacitor, as shown above, reduces the voltage drop so more KVA of load may be freed up so more load can be added without over taxing the transformer. This may be calculated by the following formula, which does not incorporate the generating or substation capacity release.

The possible increase in KW due to the released feeder KVA can be obtained by multiplying ΔKVA by the corrected power factor. Reduced Energy Losses With the decrease conductor losses due to the addition of capacitors, less kilowatt-hours of electrical energy are dissipated annually. The quantity of energy saved and the resulting economy can be computed as:

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Where

KVAR = Three phase kilovars applied. KVA = Uncorrected three-phase load (24 hour rms value…for 1200 KVA assume 720 KVA) KV = Phase to phase voltage in KV

The value of this energy savings can be obtained by multiplying the energy saved by the cost per kw-hr. All these benefits will have financial benefits to the utility, or whoever provides the capacitor bank on their system. Consolidation of Formulas 1. Power Factor Relation

Where

Kw = kilowatt load Kva = kilovolt-amperes Pf = power factor

2. Ratio of I2R Losses

3. Voltage Drop in A Line

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Where

Kva = three phase kva L = Line length in miles (1 wire only) R = Ohms resistance per mile (table 2) X = Ohms reactance per mile (table 2)

= Uncorrected power factor

= Sine of power factor angle (table 1) kv= Phase to phase kilovolts

4. Voltage Rise in A Line

Where

Kvar = three-phase kilovars applied See Formula 3 for other units

5. Voltage Rise in A Transformer

Where

Kvar = Three-phase kilovars applied KvaT = kva of transformer XT = Transformer Reactance in percent.

6. Increase in feeder Capacity

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Where

= Increase in kva capacity See Formula 3 for other units

7. Reduced Energy Losses in A Line

Where

Ea = Annual conserved energy in kw-hrs. R = Resistance to load center in Ohms. Kva = Uncorrected 3-phase kva (24-hour rms value) Kvar = here phase kilovars applied. See Formula 3 for other units

8. Increased Revenue Due to Voltage Improvement.

Where

V2 = Average voltage after adding capacitors. V1 = Average voltage before adding capacitors.

9. Reduced Substation Capacity

When kvar is small compared to kvaS

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When kvar is more than 10% of kvaS

Where

= Released kva of the substation capacity at original power factor. KvaS = Substation kva capacity Kvar = Three phase kilovars applied. See Formula 3 for other units

10. Reduced Substation Capacity When kvar is small compared to kvaG

Else

Where

= Released kva beyond maximum generating capacity at original power factor. kvaG = Generating station kva capacity